Static block in Java not executed

Question

class Test {
    public static void main(String arg[]) {    
        System.out.println("**MAIN METHOD");
        System.out.println(Mno.VAL); // SOP(9090);
        System.out.println(Mno.VAL + 100); // SOP(9190);
    }

}

class Mno {
    final static int VAL = 9090;
    static {
        System.out.println("**STATIC BLOCK OF Mno\t: " + VAL);
    }
}

I know that a static block executed when class loaded. But in this case the instance variable inside class Mno is final, because of that the static block is not executing.

Why is that so? And if I would remove the final, would it work fine?

Which memory will be allocated first, the static final variable or the static block?

If due to the final access modifier the class does not get loaded, then how can the variable get memory?

Answer 1

1. A static final int field is a compile-time constant and its value is hardcoded into the destination class without a reference to its origin;
2. therefore your main class does not trigger the loading of the class containing the field;
3. therefore the static initializer in that class is not executed.

In specific detail, the compiled bytecode corresponds to this:

public static void main(String arg[]){    
    System.out.println("**MAIN METHOD");
    System.out.println(9090)
    System.out.println(9190)
}

As soon as you remove final, it is no longer a compile-time constant and the special behavior described above does not apply. The Mno class is loaded as you expect and its static initializer executes.

Answer 2

The reason why the class is not loaded is that VAL is final AND it is initialised with a constant expression (9090). If, and only if, those two conditions are met, the constant is evaluated at compile time and "hardcoded" where needed.

To prevent the expression from being evaluated at compile time (and to make the JVM load your class), you can either:

• remove the final keyword:

  static int VAL = 9090; //not a constant variable any more
  

• or change the right hand side expression to something non constant (even if the variable is still final):

  final static int VAL = getInt(); //not a constant expression any more
  static int getInt() { return 9090; }
  

Answer 3

If you see generated bytecode using javap -v Test.class, main() comes out like:

public static void main(java.lang.String[]) throws java.lang.Exception;
    flags: ACC_PUBLIC, ACC_STATIC
    Code:
      stack=2, locals=1, args_size=1
         0: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
         3: ldc           #3                  // String **MAIN METHOD
         5: invokevirtual #4                  // Method java/io/PrintStream.println:(Ljava/lang/String;)V
         8: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
        11: sipush        9090
        14: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        17: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
        20: sipush        9190
        23: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        26: return        

You can clearly see in "11: sipush 9090" that static final value is directly used, because Mno.VAL is a compile time constant. Therefore it is not required to load Mno class. Hence static block of Mno is not executed.

You can execute the static block by manually loading Mno as below:

class Test{
    public static void main(String arg[]) throws Exception {
        System.out.println("**MAIN METHOD");
        Class.forName("Mno");                 // Load Mno
        System.out.println(Mno.VAL);
        System.out.println(Mno.VAL+100);
    }

}

class Mno{
    final static int VAL=9090;
    static{
        System.out.println("**STATIC BLOCK OF Mno\t:"+VAL);
    }
}

Answer 4

1. Actually you have not extends that Mno class so when compilation start it will generate constant of variable VAL and when execution start when that variable is needed its load thats from memory. So its not required that your class reference so that static bock is not executed.

2. if class A extends class Mno, the static block is included in class A if you do this then that static block is executed. For example..

   public class A extends Mno {
   
       public static void main(String arg[]){    
           System.out.println("**MAIN METHOD");
           System.out.println(Mno.VAL);//SOP(9090);
           System.out.println(Mno.VAL+100);//SOP(9190);
       }
   
   }
   
   class Mno {
       final static int VAL=9090;
       static {
           System.out.println("**STATIC BLOCK OF Mno\t:"+VAL);
       }
   }
   

Answer 5

As far as I know, it will be executed in order of appearance. For instance :

 public class Statique {
     public static final String value1 = init1();

     static {
         System.out.println("trace middle");
     }
     public static final String value2 = init2();


     public static String init1() {
         System.out.println("trace init1");
         return "1";
     }
     public static String init2() {
         System.out.println("trace init2");
         return "2";
     }
 }

will print

  trace init1
  trace middle
  trace init2

I just tested it and the statics are initialized (=> print) when the class "Statique" is actually used and "executed" in another piece of code (my case I did "new Statique()".